# Finding the projectile motion

1. Sep 7, 2011

### andorei

The problem goes like this.
A lady holds a hose 0.8m off the ground such that the water coming out of it reaches a point 2m away.

So the given are
a(acceleration): 9.8m/s
value for y: 0.8m
value for x: 2.0m

To find:
t (time)
Velocity for both x and y

I forgot to bring my notes so I can't clearly remember what the exact problem is but I'm sure of all the values given.

and ohh. Since I forgot my notes I can't show you guys my work.

2. Sep 7, 2011

3. Sep 7, 2011

### andorei

Wait I found the question.

Here is the exact question:
A woman holds a hose 0.8m above the ground such that the water shoots out horizontally. The water hits the ground at a point 2m away. What is the speed with which the water leaves the hose.

I have troubles finding the all the items needed.

4. Sep 7, 2011

### mstud

1. Try to solve: $x+y=v_0t + \frac 12 at^2$ for t.

2. Find t_x and t_y in the same way (you need them to find velocity...

Then find $v_x=v_0+at_x$ and $v_y=v_0+at_y$

I think this should become right

5. Sep 7, 2011

6. Sep 7, 2011

### mstud

only heighth and displacement is not enough to find start velocity... Hint: acceleration a_y=9.81 m/s^2 , A_x=0.0 m/s^2

You'll also need one more variable...

Last edited by a moderator: May 5, 2017
7. Sep 7, 2011

### andorei

What variable do i need?

thanks for the help.

8. Sep 7, 2011

### BruceW

The motion is constant acceleration, so you need the acceleration (simply gravity) and the initial position and velocity (which were given in the question). Give it a try.

9. Sep 7, 2011

### andorei

but no value of velocity was given although a direction is given to where the water shoots out.

10. Sep 7, 2011

### mstud

No, the initial velocity is what you shall find. I might think he is referring to the velocity when the water has just met the ground becomes zero?

Not quite sure if this assumption is right, however, because the water has speed when approaching the ground...

gives me,(very fastly calculated) 7.4 m/s, what about that?

Last edited: Sep 7, 2011
11. Sep 7, 2011

### andorei

What formula did you use?
I've been bugged by this question since early this morning. ugghh

12. Sep 7, 2011

### mstud

$$2a*(displacement)=v^2-(v_0)^2$$

See if you can make use of that one ...

13. Sep 7, 2011

### andorei

Wait how about I use this one to find the time?

t=$\sqrt{\frac{2d_{y}}{g}}$

will this one work then solve for the other missing values?

14. Sep 7, 2011

### BruceW

Woops, yeah sorry about that. The initial velocity is what you are trying to find.

The key to this question is that you can solve the motion in x-direction and y-direction separately, since they don't affect each other.
From the motion in y-direction, you can get the time it takes for the water to hit the ground. And then use this in the equation for the motion in x-direction to get the initial speed.

So you need to use two equations. For y-motion, it is constant acceleration, and zero initial speed (giving you time). for x-motion, use time and distance to get the (constant) speed.

15. Sep 7, 2011

### BruceW

Yep, that's exactly right.

16. Sep 7, 2011

### andorei

Soooooo, the answer is 0.16s. Is that correct? and is it also what I need to use for finding the other values?

17. Sep 7, 2011

### BruceW

I get a different answer for the time. Although your equation is correct.
Just to check: dy=0.8m, right? I think you forgot to square-root.

Once you get the time, then that is also the time which the water travels in the x-direction for. Since there is no horizontal force, the speed in the x-direction is constant. So then its just speed=dx/time